A) \[\frac{1}{{{2}^{m+n}}}\]
B) (m + n)
C) (n - m)
D) \[{{2}^{(n-m)}}\]
Correct Answer: D
Solution :
Rate becomes \[{{x}^{y}}\] times if concentration is made x times of a reactant giving \[{{y}^{th}}\] order reaction. Rate \[=k\,{{[A]}^{n}}{{[B]}^{m}}\] Concentration of A is doubled hence \[x=2,\,y=n\] and rate becomes \[={{2}^{n}}\] times Concentration of B is halved, hence \[x=1/2\] and y = m and rate becomes \[={{\left( \frac{1}{2} \right)}^{m}}\] times Net rate becomes \[={{(2)}^{n}}{{\left( \frac{1}{2} \right)}^{m}}\] times \[={{(2)}^{n-m}}\] times Alternative Solution Rate (R = K) \[{{[A]}^{n}}{{[B]}^{m}}\] ... (i) When concentration of A is doubled an concentration of B its halved, the rate becomes \[R''=k{{[2A]}^{n}}{{\left[ \frac{1}{2}B \right]}^{m}}\] \[R''=k{{2}^{n}}{{[A]}^{n}}{{\left( \frac{1}{2} \right)}^{m}}{{[B]}^{m}}\] \[R''={{2}^{n-m}}R'\]
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